Re^4: Puzzling $| behavior

i think it's matter of the compiler optimizer:
if i have something like f($x+=2, $x+=3); (and comma between arguments is not a sync point, like in C) then the optimizer could collapse $x+=2 and $x+=3 to $x+=5. It's not only matter of order.

$_ = 1;
f($_+=2, $_+=3);
sub f { print shift, ", ", shift, "\n"; }
print `perl -v`;
__________
6, 6

This is perl, v5.6.0 built for darwin
....
[download]

Oha

Comment on Re^4: Puzzling $\| behavior Download Code

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Re^5: Puzzling $\| behavior by ikegami (Patriarch) on Oct 08, 2007 at 16:58 UTC
It's not because of an optimization. `# $_ $_[0] $_[1] $_[0] $_[1] # ----- ----- ----- ----- ----- $_ = 1; # 1 do { # local @_; # alias $_[0] = $_+=2; # 3 $_ 3 alias $_[1] = $_+=3; # 6 $_ $_ 6 6 &f; };` [download] Add `$_++;` to `f` and you'll see it print `7, 7` because `$_[0]` and `$_[1]` are aliased to `$_`. It works that way because `+=` returns its LHS as an lvalue. `>perl -le "$_=1; ($_+=2)+=3; print $_" 6` [download]	[reply] [d/l] [select]

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Re^5: Puzzling $| behavior
by ikegami (Patriarch) on Oct 08, 2007 at 16:58 UTC

It's not because of an optimization.

                              # $_     $_[0]  $_[1]  $_[0]  $_[1]
                              # -----  -----  -----  -----  -----
$_ = 1;                       # 1
do {                          #
   local @_;                  #
   alias $_[0] = $_+=2;       # 3      $_            3
   alias $_[1] = $_+=3;       # 6      $_     $_     6      6
   &f;
};
[download]

Add $_++; to f and you'll see it print 7, 7 because $_[0] and $_[1] are aliased to $_.

It works that way because += returns its LHS as an lvalue.

>perl -le "$_=1; ($_+=2)+=3; print $_"
6
[download]

[reply]
[d/l]
[select]